Question 30030
Hi,

I don't know if there is an equation. I would have thought there was, but I don't know it, so I did the derivation from scratch.

I worked out the equation the really long way, by taking a general parabolic mirror, and reflecting every possible (vertical) ray that could hit it, and I solved for where they all met. I then realised there was an easier way to do it, which I'll show you, but the first technique is quite satisfying when you get it right :-)

Anyway, this is a bit simpler. Let's define that anything below the x axis is the real mirror, and anything above the x axis doesn't really exist. If we choose the center of the mirror to be at *[tex x=0] (a very wise choice, trust me!) then an equation describing the mirror would be *[tex y=k(x-d/2)(x+d/2)] where *[tex d] is the diameter and *[tex k] is some constant depending on how much we bend the mirror.

Expanding we get *[tex y=kx^2-\frac{kd^2}{4}]

Now, let's think about the point on the mirror with gradient 1. The normal at this point is at 45 deg, or 135 deg depending on how you measure. The point is that any ray that hits this part of the mirror will be reflected horizontally. It will also hit the focal point, so the y value of the focal point is the y value here!

We need to find the y value where the gradient is one, to do this we differentiate and set to one to find x. Then sub that x back in to find y.

*[tex \frac{dy}{dx}=2kx=1]
*[tex x=\frac{1}{2k}]
*[tex y=k\frac{1}{2k}^2-\frac{kd^2}{4}]
*[tex y=\frac{1}{4k}-\frac{kd^2}{4}]

The focal length is the distance from the back of the mirror(depth) to the focal point. The depth is *[tex \frac{kd^2}{4}] so *[tex y=f-\frac{kd^2}{4}]. Putting these two equations together, we can solve for *[tex k]

*[tex f-\frac{kd^2}{4}=\frac{1}{4k}-\frac{kd^2}{4}]
*[tex k=\frac{1}{4f}]

We know the depth is *[tex \frac{kd^2}{4}] so subbing for k, *[tex \frac{d^2}{16f}]

All you have to do now is put the numbers in.

Hope that helps,
Kev